The side length of a square is decreasing at a rate of $2$ kilometers per hour. At a certain instant, the side length is $9$ kilometers. What is the rate of change of the area of the square at that instant (in square kilometers per hour)? Choose 1 answer: Choose 1 answer: (Choice A) A $-81$ (Choice B) B $-4$ (Choice C) C $-324$ (Choice D) D $-36$
Setting up the math Let... $s(t)$ denote the square's side length at time $t$, and $A(t)$ denote the square's area at time $t$. We are given that $s'(t)=-2$ and that $s(t_0)=9$ for a specific time $t_0$. Note that $s'(t)$ is negative because the side length is decreasing. We want to find $A'(t_0)$. Relating the measures $A(t)$ and $s(t)$ relate to each other through the formula for the area of a square: $A(t)=[s(t)]^2$ We can differentiate both sides to find an expression for $A'(t)$ : $A'(t)=2s(t)s'(t)$ Using the information to solve Let's plug ${s(t_0)}={9}$ and ${s'(t_0)}={-2}$ into the expression for $A'(t_0)$ : $\begin{aligned} A'(t_0)&=2{s(t_0)}{s'(t_0)} \\\\ &=2({9})({-2}) \\\\ &=-36 \end{aligned}$ In conclusion, the rate of change of the area of the square at that instant is $-36$ square kilometers per hour. Since the rate of change is negative, we know that the area is decreasing.